How many oxygen atoms in a mass of $132 \cdot g$ $132*g$ with respect to carbon dioxide?

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How many oxygen atoms in a mass of $132 \cdot g$ $132*g$ with respect to carbon dioxide?

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Answer 1 · 2016-04-17T08:33:55

There are $3$ $3$ $m o l$ $mol$ of $C O_{2}$ $CO_2$ in such a mass. Thus there are $6 \times N_{A}$ $6xxN_A$ individual oxygen atoms.

Explanation:

What is $N_{A}$ $N_A$ ? It is simply $Avogadro's number$ $"Avogadro's number"$ , $6.022 \times 10^{23} \cdot m o l^{- 1}$ $6.022xx10^23*mol^-1$ .

If I have $6.022 \times 10^{23}$ $6.022xx10^23$ individual items of stuff, I have a mole of that stuff. $1$ $1$ $m o l$ $mol$ of $C O_{2}$ $CO_2$ is equivalent to a mass $44.0 \cdot g$ $44.0*g$ ; $12.01$ $12.01$ $g$ $g$ of $C$ $C$ , and $32.0$ $32.0$ $g$ $g$ of $O$ $O$ .

Answer 2 · 2016-04-17T08:40:03

$36.132 \times 10^{23}$ $36.132 times 10^23$ oxygen atoms

Explanation:

If we work out the relative formula mass of $C O_{2}$ $CO_2$

Relative atomic mass, $A_{r}$ $A_r$ of $C = 12$ $C= 12$
Relative atomic mass, $A_{r}$ $A_r$ of $O = 16$ $O=16$ , so....

$M_{r}$ $M_r$ of $C O_{2} = 44$ $CO_2 =44$

The number of moles of a substance is given by :

$\frac{mass in grams}{M_{r}}$ $"mass in grams" /(M_r)$

So in this case:

$\frac{132}{44} = 3 \cdot moles$ $132/44=3*"moles"$

One mole contains $6.022 \times 10^{23}$ $6.022times 10^23$ molecules (Avogadro's number).

So 3 moles contains $18.066 \times 10^{23}$ $18.066 times 10^23$ molecules.

Each molecule contains 2 oxygen atoms, so there will be $36.132 \times 10^{23}$ $36.132 times 10^23$ oxygen atoms

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How many oxygen atoms in a mass of $132 \cdot g$ $132*g$ with respect to carbon dioxide?

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How many oxygen atoms in a mass of 132⋅g132*g with respect to carbon dioxide?

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How many oxygen atoms in a mass of $132 \cdot g$ $132*g$ with respect to carbon dioxide?